Uniform E field for spherical shell.

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To achieve a uniform electric field within a nonconducting spherical shell with a given volume charge density ρ = A/r, the value of A must be determined based on the contributions from both the central charge and the charge density. The electric field is influenced by the central charge and the varying charge density, leading to a derived expression that combines these effects. The attempt to set the derivative of the electric field with respect to radius r to zero indicates the need for integration to account for the non-constant charge density. The user realizes that using Gauss' law directly is not feasible without first integrating to find the total charge. The discussion highlights the complexity of achieving a uniform electric field in this scenario.
GwtBc
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Homework Statement


In the figure a nonconducting spherical shell of inner radius a = 2.07 cm and outer radius b = 2.51 cm has (within its thickness) a positive volume charge density ρ = A/r, where A is a constant and r is the distance from the center of the shell. In addition, a small ball of charge q = 45.8 fC is located at that center. What value should A have if the electric field in the shell (arb) is to be uniform?

Homework Equations


## \phi = \frac{q}{\epsilon_{0}} = \oint E \cdot dA ##

The Attempt at a Solution


I found the electric field due to the central charge and ρV at a radius between r (arb), added these together (superposition) and then derived with respect to r, hoping to set the derivative to zero to find my answer, but I got ## \frac{Aa^3}{r^4}- \frac{q}{2\pi r^3} ## as my expression for the derivative, (disregarding epsilon naught, the constant) and obviously I can't set this to zero and disregard r. Not sure where to go from here.
 
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Ok, I can't use Gauss' law without integration cause charge density isn't constant, have to integrate to find charge first. So derp, I got it now.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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